{
  "id": 5098,
  "title": "HF Data",
  "url": "/competitions/belkin-energy-disaggregation-competition/discussion/5098",
  "author_name": "",
  "post_date": "2013-07-15T03:33:30.433Z",
  "votes": null,
  "comment_count": 2,
  "views": 1641,
  "content": "<p>In [4], the HF signal is sampled at 1 MHz and FFTs are computed to get 2048 frequency data points from DC to 500 kHz. In the LoaderScripts, it plots now 4096 points from DC to 1 MHz, I guess the sampling rate was upped to 2 MHz to generate this data? Could\r\n someone describe the processing that transforms the raw FFT output to unsigned 8 bit integers?</p>",
  "messages": [
    {
      "id": "27207",
      "postDate": "07/15/2013 03:33:30",
      "content": "<p>In [4], the HF signal is sampled at 1 MHz and FFTs are computed to get 2048 frequency data points from DC to 500 kHz. In the LoaderScripts, it plots now 4096 points from DC to 1 MHz, I guess the sampling rate was upped to 2 MHz to generate this data? Could\r\n someone describe the processing that transforms the raw FFT output to unsigned 8 bit integers?</p>",
      "rawMarkdown": "",
      "votes": null
    },
    {
      "id": "27213",
      "postDate": "07/15/2013 05:40:35",
      "content": "<p>You are right. In the commercial version that Belkin has built, the sampling frequency is 2Mhz. This gives us a wider spectrum and possibility to capture noise from more devices.</p>\r\n<p>The HF matrix is FFT magnitude only and are like images. Unfortunately, I cannot go into details of how we convert raw FFTs into the final uint8 matrices. That is specific to the hardware and for all practical purposes, when I analyze these matricies, I\r\n just see them as FFT magnitude spectrograms.</p>\r\n<p>Sidhant</p>",
      "rawMarkdown": "",
      "votes": null
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    {
      "id": "27925",
      "postDate": "07/31/2013 22:01:40",
      "content": "<p>Sidhant,</p>\n<p>When you say that the &quot;HF matrix is FFT magnitude&quot;, effectively, the data provided are periodogram estimates (ie naively amplitude square?) Right? Isn't the original time-series tapered before fft-ing? Shouldn't one worry about bias, especially since the data captured has many simultaneous sources, and periodogram estimate will be corrupted by close by but distinct sources?One might still get away with such an approach if one is assured that there are no 2 devices with overlapping spectral bands... but then again, given the vast array of home devices, how can such an assurance be provided...</p>\n<p>Generally speaking, using periodograms is the worst possible way to get spectral estimates...so I'm a bit confused, and concerned, that the competition provides data which is biased to begin with... Please correct me if I'm off the mark here. </p>\n<p>Thanks.</p>\n<p>&nbsp;</p>\n<p>&nbsp;</p>\n<p>&nbsp;</p>\n<p>&nbsp;</p>",
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      "votes": null
    }
  ],
  "comments": [
    {
      "id": 27213,
      "author_name": "sidhantgupta",
      "author_url": "",
      "post_date": "07/15/2013 05:40:35",
      "content": "<p>You are right. In the commercial version that Belkin has built, the sampling frequency is 2Mhz. This gives us a wider spectrum and possibility to capture noise from more devices.</p>\r\n<p>The HF matrix is FFT magnitude only and are like images. Unfortunately, I cannot go into details of how we convert raw FFTs into the final uint8 matrices. That is specific to the hardware and for all practical purposes, when I analyze these matricies, I\r\n just see them as FFT magnitude spectrograms.</p>\r\n<p>Sidhant</p>",
      "votes": null,
      "replies": []
    },
    {
      "id": 27925,
      "author_name": "nulled",
      "author_url": "",
      "post_date": "07/31/2013 22:01:40",
      "content": "<p>Sidhant,</p>\n<p>When you say that the &quot;HF matrix is FFT magnitude&quot;, effectively, the data provided are periodogram estimates (ie naively amplitude square?) Right? Isn't the original time-series tapered before fft-ing? Shouldn't one worry about bias, especially since the data captured has many simultaneous sources, and periodogram estimate will be corrupted by close by but distinct sources?One might still get away with such an approach if one is assured that there are no 2 devices with overlapping spectral bands... but then again, given the vast array of home devices, how can such an assurance be provided...</p>\n<p>Generally speaking, using periodograms is the worst possible way to get spectral estimates...so I'm a bit confused, and concerned, that the competition provides data which is biased to begin with... Please correct me if I'm off the mark here. </p>\n<p>Thanks.</p>\n<p>&nbsp;</p>\n<p>&nbsp;</p>\n<p>&nbsp;</p>\n<p>&nbsp;</p>",
      "votes": null,
      "replies": []
    }
  ],
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}